18.100B HOMEWORK 6, WAS DUE 9 MARCH 2004
Richard Melrose Department of Mathematics, Massachusetts Institute of Technology rbm@math.mit.edu
√
n2 +n−n2
we can compute the limit as
Problem 2. Since n2 + n − n = √
n2 +n+n
1
lim p
= 1.
1 + 1/n + 1
(1)
n→∞
Problem 7. By the Cauchy-Schwarz inequality,
!2
N √
N
N
X
X
an
1 X
(2)
≤
an .
n
n2 n=1
n=1
n=1
Both series on the right are convergent, hence the partial sums are bounded
so the partial sum on the left is bounded, hence, being a series of nonnegative terms, convergent.
Problem 12. (a) Since the an > 0, rn is strictly decreasing as n increases. Thus for
m < n,
(3)
am
an
1
rn − rm
rn
+ ··· +
>
(am + . . . an ) =
=1−
.
rm
rn
rm
rm
rm
P an
It follows that the series
rn is not Cauchy since the right side tends to
n
(4)
1 as n → ∞ for fixed m. Thus the series does not converge.
√
√
√
√
(b) Using the identity ( rn − rn+1 )( rn − rn+1 ) = rn − rn+1 = an
and the fact that rn is strictly decreasing, we conclude that
√ √
√
an < 2 rn ( rn − rn+1 )
giving the desired estimate. From this inequality we find that
q
X
√
√
an
√
√ < r1 − rp+1 < r1
rn
n=1
(5)
so this series with positive terms is bounded and hence convergent.
Problem 16. (a) Proceeding
inductively
1) that
√ we can assume
√ (since it is true for n =√
√
xn > α. Then x2n − 2 αxn + α = (x − α)2 > 0 so x2n + α > 2xn α and
hence
√
1
α
xn+1 =
xn +
> α.
2
xn
Since α/xn < xn also follows that√ xn+1 < xn so the sequence √
is strictly
decreasing but always larger than α. Thus the limit xn → x ≥ α exists.
√
2
2
Since 2xn xn+1 = x2n −α the
√ limit must satisfy 2x = x −α, that is x = α.
(b)Defining n = xn − α we find that
n+1 =
√
1
2
2
x2n − 2xn α + α = n < √n .
2xn
2xn
2 α
1
2
18.100B HOMEWORK 6
√
Since this is true for all n, if we set γn = n /β, where β = 2 α then
n
γn+1 < γn2 =⇒ γn+1 < γ12 ,
so n+1 < β
2 n
1
β
.
√
√
7
8
2
(c) If α = 3 and x1 = 2 then 1 10
< 3 < 1 10
so 1 = 2 − 3 < 10
,
√
−32
−16
1
˙
2 3 > 2 and 1 /β < 10 . Since β < 4, 5 < 4 · 10
and 6 < 410 .
Problem 20. Suppose thet {pn } is a Cauchy sequence and some subsequence {pn(k) }
converges to p. Then, given > 0 there exists N such that for n, m ≥ N
d(pn , pm ) < /2 and there exists N 0 such that k > N 0 implies d(p, pn(k) ) <
/2. We can choose k > N 0 so large that n(k) > N and then
d(p, pn ) ≤ d(p, pn(k) ) + d(pn , pn(k) ) < /2 + /2 = provided only that n ≥ N. Thus pn → p.
Problem 21. If {En } is a decreasing sequence of non-empty closed sets in a metric
space then there is a sequence {pn } with pn ∈ En . The assumption that
diam En → 0 means that given > 0 there exists N such that n ≥ N
implies d(p, q) < if p, q ∈ En . Now, for n ≥ m ≥ N, pn ∈ En ⊂ Em so
d(pn , pm ) < . It follows that the sequence is Cauchy and hence, by the
assumed completeness of X that it converges
to p. Since the sequence is in
T
En for m ≥ n, p ∈ En for all n so p ∈T n En as desired. Conversely there is
only one point in this set since q ∈ n En implies d(p, q) ≤ diam(En ) → 0
so p = q.